Problem Set 4 Solutions: 8.3, 8.4, 8.5, 8.6

BE.011/2.772 Problem Set 4 Solutions Due March 8, 2004 Dill 8.3, 8.4, 8.5, 8.6 8.3) The difference between the energy and enthalpy changes in expanding an ideal gas. How much heat is required to cause the quasi-static isothermal expansion of one mole of an ideal gas at T = 500 K from PA = 0.42 atm, VA = 100 L, to PB = 0.15 atm? To start with, we know that for an ideal gas in an isothermal process, dU = δq + δw = 0. For such a process, when it is also quasi-static, we also know that δw = -PdV. Therefore V nRT q = ∫ PdV . We substitute the ideal gas law for P and we get: q = ∫ dV = nRT ln A . V VB a) What is VB? PV P V 0.42 *100 By the ideal gas law, so A = B so VB = A A = = 280L PB 0.15 PB V A We can use VB to calculate the heat: V 280 q = nRT ln A = 1mol * 8.314JK - 1mol - 1 * 500Kln = 4280J VB 100 Since q is positive, heat is flowing into the system during the process. (Note that this corresponds to work being done by the system.) b) What ∆U for this process? ∆U = 0 for an isothermal process on an ideal gas since ∆U = CV∆T = 0. c) What ∆H for this process? H = U+PV so ∆H = ∆U+∆(PV) We already established that ∆U = 0. For an ideal gas at constant temperature it’s also true that PV is constant (PV = nRT). So, ∆H = 0. 8.4) The work and the heat of boiling water. For the reversible boiling of five moles of liquid water to steam at 1000C and 1 atm pressure, calculate q. Is w positive or negative? We can get the answer from tables that give the heat of vaporization of water (the answer for 1000C is given in example 8.7 but you could also use Google). Also, from the periodic table we know that water weighs 18 g/mol. qboiling = -qcondensation = 1∆Hcondensation cal g = 540 18 5moles = 48600cal g mole V2 At constant pressure, w = − ∫ PdV = − P(V2 − V1 ) = − P(V gas − Vliquid ) V1 And Vgas >> Vliquid, so Vgas – Vliquid is positive, so w is negative. Note that negative w means the system is doing work on the surroundings. 8.5) The entropy and free energy of gas expansion. Two moles of an ideal gas undergo an irreversible isothermal expansion from VA = 100 liters to VB = 300 liters at T = 300K. a) What is the entropy change for this process? dq ∆S = ∫ rev T Since S and U are state functions, we needn’t be concerned that the actual process is irreversible (read: complicated). We can choose any path to calculate state functions so we invent an isothermal, quasi-static process to get qrev. V ∆U = 0 ⇒ q rev = − wrev = nRT ln 2 . V1 For an isothermal process, q V 300 cal cal ) ln ∆S = rev = nR ln 2 = (2moles)(2 = 4.39 T V1 100 molK K b) What is the Gibbs free energy change? ∆G = ∆H − T∆S . For an isothermal process on an ideal gas, ∆H=0. (Note that ∆H = ∆U + ∆ ( PV ) and for an isothermal process, ∆U=0, and ∆ ( PV ) = ∆ (nRT ) = 0 , so ∆H=0.) cal ∆G = −T∆S = −(300K )(4.39 ) = −1318.33cal K 8.6) The free energy and entropy of membrane melting. We have a phospholipid membrane that “melts” at Tm = 410C = 314K. Reversible melting experiments indicate that ∆Hm = 9 kcal mol-1. a) What is the entropy of melting? ∆H m 9kcal * mol −1 ∆S m = = = 28.7cal * mol −1 K −1 314K T b) What is the free energy of melting? ∆G = 0 because during a reversible phase transition, the two phases are in equilibrium. Also, ∆G = ∆H - T∆S, and plugging in our numbers from above, ∆G=0. c) Does the membrane become more or less ordered on melting? ∆Sm > 0 so the entropy increases as we melt the membrane. Thus it becomes less ordered during melting. d) If there are 32 rotatable bonds in each phospholipid molecule, what is the increase in multiplicity on melting one mole of bonds? First, if there are 32 rotatable bonds in each molecule, then one mole of phospholipid corresponds to 32 moles of bonds. Thus, the entropy change per mole of bonds during melting is 28.7/32 = 0.897 cal mol-1 K-1. ln W2 ∆S 0.897 = = = 2.71e23 K 3.30e − 24 W1